论某知识点
设dp[i][j]dp[i][j]dp[i][j]表示从iii出发,往后走2j2^j2j步的最大值
然后
dp[i][j]=max(dp[i][j−1],dp[i+(1<<(j−1))][j−1]);dp[i][j]=max(dp[i][j-1],dp[i+(1<<(j-1))][j-1]); dp[i][j]=max(dp[i][j−1],dp[i+(1<<(j−1))][j−1]);
平方和公式:
n(n+1)(2n+1)6\frac {n(n+1)(2n+1)} {6} 6n(n+1)(2n+1)
立方和公式
(1+2+3+...+n)2(1+2+3+...+n)^2 (1+2+3+...+n)2